Question

Choose the correct option and justify your choice: $\sin 2A=2\sin A$ is true when $A=$

• A. $0°$
• B. $30°$
• C. $45°$
• D. $60°$

Answer 1

The correct option is A

$0°$

Determine the trigonometric values.

The explanation for the correct option:

Option (A). $0°$

Given: $\sin 2A=2\sin A$

Substitute $A=0°$ in the given equation $\sin 2A=2\sin A$ and check whether LHS is equal to RHS or not:

$\begin{array}{rcl}\mathrm{LHS}& =& \sin 2A\\ & \Rightarrow & \mathrm{LHS}=\sin \left(0°\right)\\ & \Rightarrow & \mathrm{LHS}=0\end{array}$

Now,

$\begin{array}{rcl}\mathrm{RHS}& =& 2\sin A\\ & \Rightarrow & \mathrm{RHS}=2\sin \left(0°\right)\\ & \Rightarrow & \mathrm{RHS}=2\times 0\\ & \Rightarrow & \mathrm{RHS}=0\end{array}$

Therefore, $\mathrm{LHS}=\mathrm{RHS}$.

Hence, the correct option is (A) i.e. $0°$.

The explanation for the incorrect options:

Option (B). $30°$

Substitute $A=30°$ in the given equation $\sin 2A=2\sin A$ and check whether LHS is equal to RHS or not:

$\begin{array}{rcl}\mathrm{LHS}& =& \sin 2A\\ & \Rightarrow & \mathrm{LHS}=\sin \left(60°\right)\\ & \Rightarrow & \mathrm{LHS}=\frac{\sqrt{3}}{2}\end{array}$

$\begin{array}{rcl}\mathrm{RHS}& =& 2\sin A\\ & \Rightarrow & \mathrm{RHS}=2\sin \left(30°\right)\\ & \Rightarrow & \mathrm{RHS}=2\times \frac{1}{2}\\ & \Rightarrow & \mathrm{RHS}=1\end{array}$

Therefore, $\mathrm{LHS}\ne \mathrm{RHS}$.

Hence, option (B) is incorrect.

Option (C). $45°$

Substitute $A=45°$ in the given equation $\sin 2A=2\sin A$ and check whether LHS is equal to RHS or not:

$\begin{array}{rcl}\mathrm{LHS}& =& \sin 2A\\ & \Rightarrow & \mathrm{LHS}=\sin \left(90°\right)\\ & \Rightarrow & \mathrm{LHS}=1\end{array}$

$\begin{array}{rcl}\mathrm{RHS}& =& 2\sin A\\ & \Rightarrow & \mathrm{RHS}=2\sin \left(45°\right)\\ & \Rightarrow & \mathrm{RHS}=2\times \frac{1}{\sqrt{2}}\\ & \Rightarrow & \mathrm{RHS}=\sqrt{2}\end{array}$

Therefore, $\mathrm{LHS}\ne \mathrm{RHS}$.

Hence, option (C) is incorrect.

Option (D). $60°$

Substitute $A=60°$ in the given equation $\sin 2A=2\sin A$ and check whether LHS is equal to RHS or not:

$\begin{array}{rcl}\mathrm{LHS}& =& \sin 2A\\ & \Rightarrow & \mathrm{LHS}=\sin \left(120°\right)\\ & \Rightarrow & \mathrm{LHS}=\frac{\sqrt{3}}{2}\end{array}$

Now,

$\begin{array}{rcl}\mathrm{RHS}& =& 2\sin A\\ & \Rightarrow & \mathrm{RHS}=2\sin \left(60°\right)\\ & \Rightarrow & \mathrm{RHS}=2\times \frac{\sqrt{3}}{2}\\ & \Rightarrow & \mathrm{RHS}=\sqrt{3}\end{array}$

Therefore, $\mathrm{LHS}\ne \mathrm{RHS}$.

Hence, option (D) is incorrect.

Hence, the correct option is (A) i.e. $0°$.